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初一化学上学期考试403

初一化学上学期考试403
初一化学上学期考试403

(E)乙酰乙酸甲酯

(7) 盐酸维拉帕米的体内主要代谢产物是

(A)N–去烷基化合物(B)O-去甲基化合物

(C)N–去乙基化合物(D)N–去甲基化合物

(E)S-氧化

2011 年中考全真拟试题(一)

化学试题

本试卷共六大题28小题,满分100分,考试时间90分钟。

注意事项:

⑴本试卷分试题卷和答题卡两部分,其中试题卷6页,答题卡2页;请将答案写在答题卡上每题对应的答题区域内,答在试题卷上无效。

⑵考试结束,请将本试题卷和答题卡一并上交。

⑶可能用到的相对原子质量:H~1 C~12 N~14 O~16 Cl~35.5 Ca~40 Cu~63.5 Ag~108

一、选择题(本题共16小题。每小题只有一个选项符合题意。每小题2分,共32分)1.空气中供我们呼吸的气体是

A.氧气B.氮气C.水蒸气D.稀有气体

2.人体所必需的"钙、铁、锌、硒"仅从食物中摄取不足时,可通过食品添加剂和保健药剂来予以补充,这里的"钙、铁、锌、硒"是指

A.原子B.分子C.元素D.单质

3.我们每天都和化学变化打交道,下列叙述不是化学变化的是

A.人吃进的食物逐渐被消化B.人体的呼吸作用

C.使用煤气作燃料烧饭做菜D.水在冰箱里结成冰

4.了解安全知识,可以更好地保护人身和财产安全。下列做法不正确的是A.发现燃气泄漏应立即开窗通风关阀门B.用亚硝酸钠腌渍鱼肉和蔬菜

C.霉变的大米、瓜子、花生等不能食用D.室内起火应压低身体逃离火场

5.生活中一些常见物质的pH如下图,下列四种物质中酸碱性与众不同的是A.肥皂水B.牙膏C.西瓜汁D.玉米粥

6. 下列食品中维生素C含量最丰富的是

A.牛奶B.鸡蛋C.面包D.青菜

7.下列基本的实验操作不正确的是

A.液体体积读数B.熄灭酒精灯C.检查气密性D.滴加液体

8.高铁酸钾K2FeO4是一种新型的高效水处理剂,K2FeO4中铁元素的化合价为A.-4 B.+2 C.+4 D.+6

9.下列化肥中属于氮肥的是

A.K2SO4 B.CO(NH2)2 C.KCl D.Ca3(PO4)2

10.根据右表判断,下列说法正确的是

A.铝的相对原子质量是13

B.钠离子核外有三个电子层

C.第二、三周期元素的原子从左至右最外层电子数逐渐增多

D.在化学变化中镁元素容易失去最外层2个电子形成镁离子

11.关于化学反应与能量变化的叙述不正确的是

A.在化学反应中只有燃烧才能放出热量

B.物质的缓慢氧化均放出热量

C.物质发生化学反应的同时伴随着能量变化

D.人们使用化石燃料是利用它燃烧产生的热量

12.下列实验中,固体质量变化符合右图的是

A.将铜片放入硝酸银溶液中

B.将饱和氢氧化钙溶液加热

C.将氯酸钾和二氧化锰混合加热

D.将碳和氧化铜混合在高温下反应

13.不需加任何试剂就能鉴别出来的一组物质是

A.氯化钠、硝酸铵、氯化铁三种固体 B. 氢气、甲烷、一氧化碳三种气体C.氯化铵、氢氧化钾、硝酸钠三种溶液 D. 酒精、汽油、蒸馏水三种液体14.下列每组对比实验不能达到实验目的是

选项

实验设计

实验目的

实验一

实验二

A

将带火星的木条伸入装有5mL5%的过氧化氢溶液的试管中

将带火星的木条伸入装有少量红砖粉和5mL5%的过氧化氢溶液的试管中

催化剂对化学反应速率的影响

B

将铁钉放入干燥的试管中并将试管口密封

将铁钉放入装有水的试管中,使铁钉一半被水淹没

铁生锈是水和氧气共同作用的结果

C

将铁丝放入2mL10%的硫酸铜溶液中

将银丝放入2mL10%的硫酸铜溶液中

探究铁、铜、银三种金属的活动性顺序

D

将植物嫩叶浸泡在10mL蒸馏水中

取10mL蒸馏水,通入过量二氧化硫气体,再将植物嫩叶浸泡其中

酸雨危害的模拟实验

15.除去下列物质中的杂质,所用的试剂和方法不正确的是

A.除去二氧化碳中混有的一氧化碳可以用点燃的方法

B.除去氯化钾中混有的二氧化锰可以用溶解、过滤、蒸发的方法

C.除去铜粉中混有的锌粉可以用加入过量的稀盐酸、过滤、洗涤、烘干的方法D.除去生石灰中混有的碳酸钙可以用高温煅烧的方法

16.小明同学对所学部分化学知识归纳如下,其中完全正确的一组是

A.物质的分类

B.资源的利用和保护

①合成橡胶、合金都属于合成材料

②氧化物中含有氧元素

③混合物中一定含有多种元素

①保护空气要减少有害气体和烟尘的排放

②爱护水资源要节约用水和防止水体污染

③保护金属资源只能通过防止金属腐蚀实现

C."低碳经济"的措施

D.用"化学"眼光分析

①改造或淘汰高能耗、高污染产业

②研制和开发新能源替代传统能源

③减少使用一次性塑料方便袋

①洗涤剂去油污是因为起乳化作用

②湿衣服晾干是因为分子之间有间隔

③明矾净水是因为明矾溶于水生成胶状物

二、填空题(共3小题。化学方程式每个3分,其余每空1分,共12分)

17.(3分)用化学符号填空:1个硫原子;2个氮分子,我发现若化学符号前面的化学计量数大于1,则该符号只能表示(填"宏观"或"微观")含义。

18.(3分)运用溶解度表与溶解度曲线回答下列问题:

⑴B是的溶解度曲线。

⑵60℃时,100g水中加入100g硝酸钾,所得溶液是(填"饱和"或"不饱和")溶液,当60℃的该溶液降温到30℃时,析出晶体的质量为g。

19.(6分)按下列要求书写化学方程式:

⑴天然气的主要成分甲烷在空气中燃烧;

⑵在通电条件下只有两种气体生成的分解反应。

三、简答(说明)题(共4小题。化学方程式每个3分,其余每空1分,共13分)

20.(3分)根据物质结构、性质、用途之间的关系,回答下列问题:

⑴物质的结构决定物质的性质。如:活性炭具有疏松多孔的结构,因此活性炭具有

性。

⑵物质的性质决定物质的用途。如:。考虑物质的用途时除了考虑物质的性质外,还要考虑多种因素,如:。

21.(4分)下图为氧化汞受热分解时的微观过程示意图:

根据以上示意图,写出该反应的化学方程式。从图示中我还发现构成物质的粒子有。

22.(4分)金属活动性顺序在运用中给我们三条判断依据,即在金属活动性顺序里:

⑴金属的位置越靠前它的活动性就越强;

⑵位于氢前面的金属能置换出盐酸、稀硫酸中的氢;

⑶位于前面的金属能把位于后面的金属从它们化合物的溶液里置换出来。

通过查阅资料我发现金属铝与氧化铜在高温条件下能发生置换反应,其化学方程式为,由此我发现了一条新的判断依据。

23.(2分)根据下列实验图示分析问题:

硫的燃烧镁带的燃烧铁丝的燃烧

⑴上述三个反应都是物质与氧发生的反应,它们都属于氧化反应。氧气在氧化反应中提供氧,它具有氧化性。我能从化学反应前后氧元素化合价变化分析,氧气在上述反应中具有氧化性的原因是。

⑵通过类比、分析反应2Na+Cl22NaCl发现,该反应中具有氧化性的物质是

(填化学式)。

四、简答(分析)题(共2小题。化学方程式每个3分,其余每空1分,共23分)24.(15分)通过一年的化学学习,我们知道了常见气体的实验室制取方法和相关性质,我能根据所学知识回答下列问题:

⑴图24-1是实验室常用的制取气体的装置:

实验室用高锰酸钾制取氧气的化学方程式是,所选用的发生装置是(填序号),用C装置收集氧气后,验满的方法是。

⑵小丽欲使用图24-2装置收集较纯净的氧气,则应先将装置中装满

水,再将气体从(填"a"或"b")端通入。

⑶小丽用排新制的澄清石灰水的方法收集了半瓶氧气,然后将燃着的

木炭伸入该集气瓶中继续做气体性质实验,我能观察到的现象是,

我能写出二氧化碳和澄清石灰水发生反应的化学方程式。

⑷炼铁的原理是利用一氧化碳与氧

化铁反应。在实验室里,可用图24-3

所示装置进行实验:实验中玻璃管内

的粉末由红色变为色,该反

应的化学方程式为,通过实

验我可以得出的一氧化碳的相关化

学性质是。

25.(8分)右图是物质化学性质的知识网络图,"-"表示相连的两种物质能发生反应,"→"表示一种物质转化成另一种物质。

⑴我能用单质、氧化物、酸、碱、盐五种不同类别物质的化学式填空,完善这个知识网络。

⑵我能写出由C→D反应的化学方程式。

五、简答(探究)题(共1小题。化学方程式每个3分,其余每空1分,共13分)26.(13分)在高效复习课堂中,同学们利用"硫酸、硝酸钡、氢氧化钠、碳酸钾" 四种物质的溶液对"复分解反应发生的条件"进行再探究。

【知识回放】

通过判断上述四种溶液两两之间能否发生反应,我能写出其中符合下列条件的两个化学方程式:

⑴一个中和反应;

⑵一个有沉淀生成的反应。

【挖掘实质】

⑴图26-1是HCl和NaOH在水中解离出离子的示意图,小丽发现HCl在水中会解离出H+和Cl-,NaOH在水中会解离出Na+和OH-,以此类推硝酸钡在水中会解离出的离子是(填离子符号)。

⑵图26-2是稀盐酸和氢氧化钠溶液发生化学反应的微观过程示意图,该化学反应的实质可以表示为

H+ + OH-=H2O

我也能用同样的方式表示上述四种溶液两两之间发生反应时有气体生成的反应的实质。

【实践运用】

⑴通过上述探究发现,在溶液中H+和OH-不能大量共存,我还发现在溶液中

和(填离子符号,下同)也不能大量共存。

⑵利用上述四种物质的溶液完成实验后,将所得物质混合并过滤,得到一种无色溶液。经测定,该溶液显酸性。则在该溶液中一定含有的离子是。为了验证该溶液中可能含有的离子是否存在,我设计如下实验进行验证。

溶液中可能含有的离子

设计实验验证(实验步骤、现象及对应的结论)

【知识升华】

对"复分解反应发生的条件"的再探究,我发现在分析多种溶液混合后所得溶液中离子的成分时,应考虑的因素是。

六、计算题(共2小题。第27小题2分,第28小题5分,共7分)

27.(2分)甲醛是室内装修的隐形杀手,甲醛的化学式为CH2O。

⑴甲醛中碳、氢、氧的质量比为;⑵甲醛中碳元素的质量分数为。28.(5分)鸡蛋壳的主要成分是碳酸钙,为了测定鸡蛋壳中碳酸钙的含量,小丽称取

30g干燥的碎鸡蛋壳放入烧杯中,并向其中加入了80g稀盐酸恰好完全反应(假设鸡蛋壳中除碳酸钙外的其他成分都不溶于水,且不与稀盐酸反应),反应后烧杯中物质的总质量为101.2g。我能完成下列计算:

⑴碳酸钙的质量;

⑵当碳酸钙恰好完全反应时所得溶液中溶质的质量分数。(结果保留一位小数)

2011 年中考模拟题

化学参考答案及评分标准

说明:本卷中开放性题的解答,只要符合题意要求,均按对应题标准给分。

一、选择题(本题共16小题。每小题只有一个选项符合题意。每小题2分;共32分)

题号

1

2

3

4

5

6

8

9

10

11

12

13

14

15

16

答案

A

C

D

B

C

D

B

D

B

C

A

A

D

B

A

C

二、填空题(共3小题。化学方程式每个3分,其余每空1分,共12分)

17.(3分)S;2N2;微观

18.(3分)⑴氯化钠或NaCl;⑵不饱和;54.2

19.(6分)⑴CH4+2O2CO2+2H2O ⑵2H2O2H2↑+O2↑(化学方程式中化学式正确得2分,配平、条件、状态标注正确得1分,下同)

三.简答(说明)题(共4小题。化学方程式每个3分,其余每空1分,共13分)

20.(3分)⑴吸附;(错别字不给分)

⑵石墨具有导电性,常用做电极等(其他合理答案也可得分,下同);价格或是否美观等。

21.(4分)2HgO 2Hg+O2↑;分子、原子(两个答全得1分,其他情况不得分)22.(4分)2Al+3CuO 3Cu+Al2O3;在高温条件下,位于前面的金属能把位于后面的金属从它们的氧化物里置换出来。

23.(2分)⑴化学反应前后氧元素的化合价降低或氧元素的化合价降低;⑵Cl2

四、简答(分析)题(共2小题。化学方程式每个3分,其余每空1分,共23分)24.(15分)⑴2KMnO4 K2MnO4 + MnO2 + O2↑;A;将带火星的木条放在集气瓶口内沿处(伸入瓶中不得分),若木条复燃,则氧气已经收集满了。⑵b

⑶木炭燃烧发出白光,放热,澄清石灰水表面、液面上集气瓶内壁出现一层白

(答全得1分,答澄清石灰水变浑浊不得分)

⑷黑;3CO+Fe2O3 2 Fe+3CO2;可燃性,还原性,毒性(三条答全得1分)

25.(8分)(评分标准:1、每条线或单箭头正确得1分,双箭头全对得1分。2、其它答案合理给分)

2HCl+Mg=MgCl2+H2↑2HCl+CuO=CuCl2+H2O

五、简答(探究)题(共1小题, 化学方程式每个3分,其余每空1分,共13分)26.(13分)

【知识回放】⑴H2SO4+2NaOH=Na2SO4+2H2O

⑵H2SO4+Ba(NO3)2=BaSO4↓+2HNO3或K2CO3+Ba(NO3)2=BaCO3↓+2KNO3 【挖掘实质】Ba2+和NO3-或Ba2+和2NO3-;2H++CO32-=H2O+CO2↑或2H++CO32-=H2CO3,H2CO3=H2O+CO2↑(若未配平,生成物未标注箭头,不扣分)

【实践运用】Ba2+和SO42-或Ba2+和CO32-或H+和CO32-;H+、Na+、K+和NO3-(必须答全四种得1分)

溶液中可能含有的离子

设计实验验证(实验步骤、现象及对应的结论)

可能性一:Ba2+(同时答两种或两种以上离子的不得分,下同。)

取上述溶液少量于试管中,加入少量硫酸钠溶液。若出现白色沉淀,则说明上述溶液中含有Ba2+;若没有出现白色沉淀,则说明上述溶液中不含有Ba2+。

可能性二:SO42-

取上述溶液少量于试管中,加入少量硝酸钡溶液。若出现白色沉淀,则说明上述溶液中含有SO42-;若没有出现白色沉淀,则说明上述溶液中不含有SO42-。

可能性三:Ba2+或SO42-

取上述溶液少量于试管中,加入少量硫酸钠溶液。若出现白色沉淀,则说明上述溶液中含有Ba2+,不含有SO42-。(若没有出现白色沉淀,必须验证SO42-)(其他答案合理给分。)【知识升华】

溶液中的离子能否大量共存或者离子是否参加化学反应,参加化学反应的离子是否有剩余。(意义表达准确,都可给分。)

六、计算题(共2小题,第27小题2分,第28小题5分,共7分)

27.(2分)⑴12︰2︰16或6︰1︰8 ⑵40%

28.(5分)解:根据质量守恒定律,生成二氧化碳的质量为80g+30g-101.2g=8.8g..................1分

设30g鸡蛋壳中碳酸钙的质量为,生成氯化钙的质量为。

CaCO3+2HCl=CaCl2+H2O+CO2↑..............................1分

100 111 44

8.8g

....................................1分

所得溶液中溶质的质量分数=...... 1分

答:⑴30g鸡蛋壳中碳酸钙的质量为20g。⑵所得溶液中溶质的质量分数为24.3%。

(解、设、答不全扣1分,三项不另设分数。)

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2、报名参与,报名邮箱:123国学网官方邮箱。谢绝空降。报名信息包含第一姓名、年龄、性别;第二手机号码、QQ号码、邮箱;第三、学习目标;第四子女孩子及家庭情况简介。

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